Bài 70 trang 40 SGK Toán 9 tập 1

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\(\displaystyle a)\sqrt {{{25} \over {81}}.{{16} \over {49}}.{{196} \over 9}}\)                            

\(\displaystyle b)\sqrt {3{1 \over {16}}.2{{14} \over {25}}2{{34} \over {81}}}\)

\(\displaystyle c){{\sqrt {640} .\sqrt {34,3} } \over {\sqrt {567} }}\)                                    

\(d)\sqrt {21,6} .\sqrt {810.} \sqrt {{{11}^2} - {5^2}}\) 

Lời giải

a) \(\eqalign& \sqrt {{{25} \over {81}}.{{16} \over {49}}.{{196} \over 9}} \cr & = \sqrt {{{25} \over {81}}} .\sqrt {{{16} \over {49}}} .\sqrt {{{196} \over 9}} \cr & = {5 \over 9}.{4 \over 7}.{{14} \over 3} = {{40} \over {27}} \cr} \)

b) \(\eqalign& \sqrt {3{1 \over {16}}.2{{14} \over {25}}2{{34} \over {81}}} \cr 

& = \sqrt {{{49} \over {16}}.{{64} \over {25}}.{{196} \over {81}}} \cr
& = \sqrt {{{49} \over {16}}} .\sqrt {{{64} \over {25}}} .\sqrt {{{196} \over {81}}} \cr
& = {7 \over 4}.{8 \over 5}.{{14} \over 9} = {{196} \over {45}} \cr} \)

c\(\begin{array}{l}\dfrac{{\sqrt {640} .\sqrt {34,3} }}{{\sqrt {567} }} = \sqrt {\dfrac{{640.34,3}}{{567}}}  = \sqrt {\dfrac{{64.49.7}}{{81.7}}} \\

 = \sqrt {\dfrac{{64.49}}{{81}}}  = \dfrac{{\sqrt {64} .\sqrt {49} }}{{\sqrt {81} }} = \dfrac{{8.7}}{9} = \dfrac{{56}}{9}
\end{array}\)

d) \(\eqalign{

& \sqrt {21,6} .\sqrt {810.} \sqrt {{{11}^2} - {5^2}} \cr
& = \sqrt {21,6.810.\left( {{{11}^2} - {5^2}} \right)} \cr
& = \sqrt {216.81.\left( {11 + 5} \right)\left( {11 - 5} \right)} \cr
& = \sqrt {{36.6}{{.9}^2}{{.4}^2}.6} = 6.9.4.6 = 1296 \cr} \)


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